Probability of player hard 20 losing to a 5 card delaer 21

This was a thread yesterday but I cannot find it, so here it is again. There was some progress made on it, but then it "died" I think.

Double deck, dealer hits soft 17.

First hand (solo player) gets two tens and dealer has an ace up. Player stands. Dealer flips over the hole card (cannot be 7-10 of course), draws a third card, draws a fourth card, and draws a fifth card that is exactly 21. You lose! Darn!

I started off with all 10,000 different way for the dealer to expose the hole card and the next three cards (4 cards having one of 10 values is 10^4=10,000).

This part I did on a spreadsheet. I started with AAAAA, then AAAA2, AAAA3, etc., all the way to ATTTT.

Then, starting at the bottom, I deleted all the ones that start with AT since that is dealer blackjack. Then I deleted more, starting from the bottom, as those would be A9xyz (soft 20 with no more cards), same for A8xyz, and A7xyz. Now my last row is A6TTT. That one is three cards ending in hard 17 (the fourth and fifth cards are irrelevant but part of the entire set of A+ 4 cards of any value). Delete A6Txy. Now my last row is A69Tx, which is bust.

At this point I decided to write a VBA (Visual Basic for Access) program to analyze the 590 possibilites. I don't like Excel Visual Basic. Access can handle millions of rows in a table (limit 2GB of data in a table, something that would bog down Excel for an eternity).

Anyway, I wrote the program, acting like the dealer, where I start with A (score of 11), flip over the hole card, add its score (if A, it must be 1 since I already used 11), and so on. If a soft score>21, switch to hard total and subtract 10 (e.g. A3T4x is hard 18, no 5th card needed).

For each row, the program output the result, which is bust, 3 card 21 (e.g. A64), 4 card 21, 3 cards soft 18-20, 3 cards hard 17-20, 4 cards soft 18-20, and 4 cards soft 17-20. What is left is 4 cards that add up to soft 17 or 16 or less. Now I can easily calculate what the 5th card must be (e.g. soft 17 on 4 cards, draw a 4 to get to 21 in 5 cards).

I ended up with 135 possibilities: (sorry for the bad formatting; haven't figured that out yet)

up card hole card third card fourth card 5th card
A A A A 7
A A A 2 6
A A A 3 5
A A A 4 4
A A A 9 9
A A A 10 8
A A 2 A 6
A A 2 2 5
A A 2 3 4
A A 2 8 9
A A 2 9 8
A A 2 10 7
A A 3 A 5
A A 3 2 4
A A 3 7 9
A A 3 8 8
A A 3 9 7
A A 3 10 6
A A 4 A 4
A A 4 6 9
A A 4 7 8
A A 4 8 7
A A 4 9 6
A A 4 10 5
A A 5 5 9
A A 5 6 8
A A 5 7 7
A A 5 8 6
A A 5 9 5
A A 10 A 8
A A 10 2 7
A A 10 3 6
A A 10 4 5
A 2 A A 6
A 2 A 2 5
A 2 A 3 4
A 2 A 8 9
A 2 A 9 8
A 2 A 10 7
A 2 2 A 5
A 2 2 2 4
A 2 2 7 9
A 2 2 8 8
A 2 2 9 7
A 2 2 10 6
A 2 3 A 4
A 2 3 6 9
A 2 3 7 8
A 2 3 8 7
A 2 3 9 6
A 2 3 10 5
A 2 4 5 9
A 2 4 6 8
A 2 4 7 7
A 2 4 8 6
A 2 4 9 5
A 2 9 A 8
A 2 9 2 7
A 2 9 3 6
A 2 9 4 5
A 2 10 A 7
A 2 10 2 6
A 2 10 3 5
A 3 A A 5
A 3 A 2 4
A 3 A 7 9
A 3 A 8 8
A 3 A 9 7
A 3 A 10 6
A 3 2 A 4
A 3 2 6 9
A 3 2 7 8
A 3 2 8 7
A 3 2 9 6
A 3 2 10 5
A 3 3 5 9
A 3 3 6 8
A 3 3 7 7
A 3 3 8 6
A 3 3 9 5
A 3 8 A 8
A 3 8 2 7
A 3 8 3 6
A 3 8 4 5
A 3 9 A 7
A 3 9 2 6
A 3 9 3 5
A 3 10 A 6
A 3 10 2 5
A 4 A A 4
A 4 A 6 9
A 4 A 7 8
A 4 A 8 7
A 4 A 9 6
A 4 A 10 5
A 4 2 5 9
A 4 2 6 8
A 4 2 7 7
A 4 2 8 6
A 4 2 9 5
A 4 7 A 8
A 4 7 2 7
A 4 7 3 6
A 4 7 4 5
A 4 8 A 7
A 4 8 2 6
A 4 8 3 5
A 4 9 A 6
A 4 9 2 5
A 4 10 A 5
A 5 A 5 9
A 5 A 6 8
A 5 A 7 7
A 5 A 8 6
A 5 A 9 5
A 5 6 A 8
A 5 6 2 7
A 5 6 3 6
A 5 6 4 5
A 5 7 A 7
A 5 7 2 6
A 5 7 3 5
A 5 8 A 6
A 5 8 2 5
A 5 9 A 5
A 6 5 A 8
A 6 5 2 7
A 6 5 3 6
A 6 5 4 5
A 6 6 A 7
A 6 6 2 6
A 6 6 3 5
A 6 7 A 6
A 6 7 2 5
A 6 8 A 5

Back to Excel, I calculated the probability of each occurrence. I have 4 tables of deck compositions, starting with the same for everyone, 30 tens, 7 aces, and 8 of everything else. The 2nd table has the deck composition after the hole card, and so on, until the last table with the deck composition after the 4th card has been drawn (to calculate the probability of the 5th needed card being drawn). I calculated the probability of that particular hand happening (e.g. AA23T5 as 1.13851693510761E-06).

Add them all together, and I get 5.00105E-05, or 1 in 19995.78.

Fun! :)

This is the thread you're referring to. Good work on the calculation. I suggest you go there and re-post your work, as I can't combine this thread to that one.

This thread is closed.

See:https://wizardofvegas.com/forum/questions-and-answers/math/31103-blackjack-5-card-charlie-probability-question/2/#post659609